y=xsin(2x), 2(1+2x²)y=2y+4x²y=2xsin(2x)+4x³sin(2x) ➀;
y'=2xcos2x+sin(2x), -2xy'=-4x²cos(2x)-2xsin(2x) ➁;
y"=-4xsin(x)+4cos(2x), x²y"=-4x³sin(x)+4x²cos(2x) ➂.
➀+➁+➂=0 QED
This can also be proved by starting at the DE:
Let u=y/x, then u'=-y/x²+y'/x, u"=2y/x³-2y'/x²+y"/x.
Therefore, y=ux, x²u'=-y+xy', x³u"=2y-2xy'+x²y".
x²y"-2xy'+2(1+2x²)y=x²y"-2xy'+2y+4x²y=x³u"+4x³u=0.
So, u"+4u=0, which has the general solution u=Asin(2x)+Bcos(2x), that is, y/x=Asin(2x)+Bcos(2x), where A and B are constants.
y=Axsin(2x)+Bxcos(2x), which has the particular solution A=1 and B=0, y=xsin(2x).